SPOJSUBST1-New Distinct Substrings

SUBST1 - New Distinct Substrings

#suffix-array-8

Given a string, we need to find the total number of its distinct substrings.

Input

T- number of test cases. T<=20; Each test case consists of one string, whose length is <= 50000

Output

For each test case output one number saying the number of distinct substrings.

Example

Input:
2
CCCCC
ABABA

Output:
5
9

题意：给定一个字符串，求不同子串的数目

解题思路：后缀数组

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const int N = 200010;

struct Sa
{
	char s[N];
	int rk[2][N], sa[N], h[N], w[N], now, n;
	int rmq[N][20], lg[N], bel[N];

	void GetS()
	{
		scanf("%s", s + 1);
		n = strlen(s + 1);
	}

	void getsa(int z, int &m)
	{
		int x = now, y = now ^= 1;
		for (int i = 1; i <= z; i++) rk[y][i] = n - i + 1;
		for (int i = 1, j = z; i <= n; i++)
			if (sa[i] > z) rk[y][++j] = sa[i] - z;
		for (int i = 1; i <= m; i++) w[i] = 0;
		for (int i = 1; i <= n; i++) w[rk[x][rk[y][i]]]++;
		for (int i = 1; i <= m; i++) w[i] += w[i - 1];
		for (int i = n; i >= 1; i--) sa[w[rk[x][rk[y][i]]]--] = rk[y][i];
		for (int i = m = 1; i <= n; i++)
		{
			int *a = rk[x] + sa[i], *b = rk[x] + sa[i - 1];
			rk[y][sa[i]] = *a == *b&&*(a + z) == *(b + z) ? m - 1 : m++;
		}
	}

	void getsa(int m)
	{
		now = rk[1][0] = sa[0] = s[0] = 0;
		for (int i = 1; i <= m; i++) w[i] = 0;
		for (int i = 1; i <= n; i++) w[s[i]]++;
		for (int i = 1; i <= m; i++) rk[1][i] = rk[1][i - 1] + (bool)w[i];
		for (int i = 1; i <= m; i++) w[i] += w[i - 1];
		for (int i = 1; i <= n; i++) rk[0][i] = rk[1][s[i]];
		for (int i = 1; i <= n; i++) sa[w[s[i]]--] = i;
		rk[1][n + 1] = rk[0][n + 1] = 0;
		for (int x = 1, y = rk[1][m]; x <= n&&y <= n; x <<= 1) getsa(x, y);
		for (int i = 1, j = 0; i <= n; h[rk[now][i++]] = j ? j-- : j)
		{
			if (rk[now][i] == 1) continue;
			int k = n - max(sa[rk[now][i] - 1], i);
			while (j <= k&&s[sa[rk[now][i] - 1] + j] == s[i + j]) ++j;
		}
	}

	void getrmq()
	{
		h[n + 1] = h[1] = lg[1] = 0;
		for (int i = 2; i <= n; i++)
			rmq[i][0] = h[i], lg[i] = lg[i >> 1] + 1;
		for (int i = 1; (1 << i) <= n; i++)
		{
			for (int j = 2; j <= n; j++)
			{
				if (j + (1 << i) > n + 1) break;
				rmq[j][i] = min(rmq[j][i - 1], rmq[j + (1 << i - 1)][i - 1]);
			}
		}
	}

	int lcp(int x, int y)
	{
		int l = min(rk[now][x], rk[now][y]) + 1, r = max(rk[now][x], rk[now][y]);
		return min(rmq[l][lg[r - l + 1]], rmq[r - (1 << lg[r - l + 1]) + 1][lg[r - l + 1]]);
	}

	void work()
	{
		getsa(300);
		int ans = n + 1 - sa[1];
		for (int i = 2; i <= n; i++)
		{
			if (h[i] <= 0) ans += n + 1 - sa[i];
			else ans += n + 1 - sa[i] - h[i];
		}
		printf("%d\n", ans);
	}
}sa;

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		sa.GetS();
		sa.work();
	}
	return 0;
}